Question 1062214
{{{ 2(x+2)^2 - 4 = 28 }}}
{{{ 2(x^2+4x+4) - 4 - 28 = 0 }}}
{{{  2x^2 + 8x + 8 - 4 -28 = 0 }}}
{{{  2x^2 + 8x - 24 = 0 }}}
{{{   x^2 + 4x - 12 = 0 }}}   (divided previous line by 2, both sides)
{{{  (x-2)(x+6) = 0 }}}

x=2 and x=-6
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Check:  
x=2:   2(x+2)^2 - 4 = 2(4)^2 - 4 = 2(16) - 4 = 32-4 = 28 (ok)
x=-6  2(x+2)^2 - 4= 2(-6+2)^2 - 4 = 2(16)- 4 = 32-4 = 28 (ok)
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Answer:
x = 2   and
x = -6  both solve the equation
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